hey guys welcome to IGCSE study Bud where you can revise chemistry topics from the Cambridge icsc syllabus if you are enjoying our video so far please don't forget to hit the like button and subscribe to our channel in this video you are going to learn part four of topic three stochiometry the mass of one mole of a substance is the molar mass or capital M so for an element the molar mass is the same as the element's relative atomic mass and in the case of a compound the molar mass is the relative formula mass or relative molecular mass molar mass is expressed in G per mole the amount of substance is equal to mass divided by molar mass we can use this relationship for various calculations let's do some calculations as examples what is the mass of 0.75 moles of sulfur so we substitute the number of moles and the molar mass in this case the relative atomic mass of sulfur to the formula amount of substance equals to mass over molar mass so mass is equal to 0.75 multiplied by 32 because we make Mass the subject of the formula by multiplying both sides by 32 this is equal to 24 G of sulfur next example how many moles are in 10.2 G of calcium carbonate ca3 in the previous example we were asked to find the mass in this example we need to find the moles in the previous example since it was an element we were dealing with the molar mass was the same as its relative atomic mass in this example we are calculating the number of moles in a compound so the molar mass will be the relative molecular mass so we have to add up the atomic masses of all the atoms in calcium carbonate moles is equal to 10.2 divided by 40 + 12 + 16 * 3 because there are three oxygen atoms in ca3 the relative molecular mass is 100 so 10.2 ided 100 is 0.102 moles another example in 10 G of methane how many molecules are there and how many atoms are there so in a question like this the first step would be to find out how many moles are there in this mass of methane number of moles equal to 10 / 12 + 1 * 4 since there are 4 hydrogen atoms that will be 0.625 moles we learned the avocados constant which is one mole of any substance contains 6 02 into 10 ^ 23 particles we just found that 10 G of methane contains 0.625 moles so if 1 mole of CH4 has 6.02 into 10 ^ 23 molecules how many molecules let's call it X are there in 0.625 moles so let's do a simple cross multiplication 1 * X is x = 0.625 * 6.02 into 10 ^ 23 that gives us 3.76 into 10 ^ 23 molecules now how do we find out how many atoms are there in CH4 one molecule of CH4 has five atoms because there's one carbon atom and four hydrogen atoms we just found that there are 3.76 into 10 ^ 23 molecules of CH4 so how many atoms does this translate to we simply multiply this by five that gives us 1.88 into 10 ^ 24 atoms moving on to calculating reacting masses here we determine the masses of reactants consumed and products formed in a chemical reaction example one calculate the mass of calcium oxide CAO formed when 6 G of calcium reacts with excess oxygen this is the balanced equation we are given the mass of calcium and we are asked to find the mass of calcium oxide produced used the mole ratio is 2 is 2 simplified it becomes 1 is 1 so we first find the number of moles in 6 G of calcium applying the mole formula we get 0.15 moles the mole ratio of calcium to calcium oxide is 1 is to 1 according to the balanced equation so that means 0 .15 mol of calcium will produce 0.15 mol of calcium oxide so 0.15 mol of calcium oxide is substituted into the mole formula to find the mass of calcium oxide produced the molar mass of CAO is 40 + 16 the mass of calcium oxide produced is is therefore 8.4 G example two calculate the mass of ion in kilog that can be produced from 250 kg of ion 3 oxide the balanced chemical equation for the reaction is 2 Fe2O3 to give 4 Fe and 32 so here the mole ratio is for every 2 moles of ion 3 oxide 4 moles of ion are produced when you simplify you get a ratio of 1 is to 2 let's first find the number of moles of fe203 keep in mind that in the mole formula the mass has to be in G so we convert 200 50 kg to G which is 250,000 G the molar mass of Fe2O3 is 56 * 2 + 16 * 3 because there are two atoms of ion and three atoms of oxygen in Fe2O3 the answer is 1, 15625 moles of ion 3 oxide we multiply this by two to get the number of moles of ion produced which is 3,125 now we substitute this number of moles into the mole formula to get 175,000 G of ion we convert this to kilog because the question requires the answer answer to be given in kilog so 175 kg of iron are produced that concludes part four of topic three stochiometry are you enjoying our videos are they helping you here's a way you can show your appreciation and support our continued efforts you may use YouTube super thanks to send us thanks hope this video helped you please share your thoughts and suggestions in the comment section thank you for watching and please don't forget to subscribe to igcc studybody for more revision videos 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